Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
//Coins
import java.io.*;
import java.util.*;
public class Main{
	static int n;
	static double[] prob=new double[3001];
	static double[][] dp=new double[3001][3001];
	public static double solve(int coins,int heads){
		if(heads==0)
			return 1;
		if(coins==0)
			return 0;
		//head comes
		if(dp[coins][heads]!=-1)
			return dp[coins][heads];
		double op1=prob[coins]*solve(coins-1,heads-1);
		//tail comes
		double op2=(1-prob[coins])*solve(coins-1,heads);
		return dp[coins][heads]=op1+op2;

	}
	public static void main(String[] args) throws IOException{
    	 BufferedReader br = new BufferedReader( 
                              new InputStreamReader(System.in)); 
       
		n=Integer.parseInt(br.readLine()); 
		// prob=new double[n+1];
		// dp=new double[]
		String[] str=br.readLine().split(" ");
		for(int i=1;i<=n;i++){			
		   prob[i]=Double.parseDouble(str[i-1]);		 
		}
		for(double[] x:dp){
			Arrays.fill(x,-1);
		}
		System.out.println(solve(n,(n+1)/2));
	
    }
}